Physics Question Over 2 Dimensional Friction Problems

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SUMMARY

The discussion centers on a physics problem involving a 2 kg bucket and a 0.5 kg glass being spun in a vertical circle with a radius of 1.5 m. Participants analyze the forces acting on the glass, including weight, normal force, and centripetal force, to determine the minimum speed required to keep the glass from falling out and the maximum tension the string can withstand without breaking. The correct equations derived include the net force equation and the relationship between centripetal force and tension, leading to solutions of approximately 3.83 m/s for part c and 9.47 m/s for part d.

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  • #31
haruspex said:
The glass is blissfully unaware of the tension in the string and of gravity acting on the bucket. Which forces act directly on the glass, and by doing so provide the centripetal force ( so don't answer "centripetal force").

Normal Force?
 
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  • #32
Xazerd said:
Thank You, but I already solved that one now ( I got Net Force= Normal Force- Weight + (Mass*Velocity^2)/radius.)
That equation is not correct, in a couple of ways. Perhaps you've typed it out wrongly?
 
  • #33
pgardn said:
T + mg = (mv^2)/r
Think again. At what point in the loop is the tension greatest?
 
  • #34
haruspex said:
Think again. At what point in the loop is the tension greatest?

Aha again!

I was using the FBD for part a)

The problem said Maximum tension. This would happen at the bottom of the loop.

So T - mg = (mv^2)/r

It always helps to answer the questions that are actually asked.
Sorry to the OP. Thanks to Mr. H!
 
  • #35
pgardn said:
Aha again!

I was using the FBD for part a)

The problem said Maximum tension. This would happen at the bottom of the loop.

So T - mg = (mv^2)/r

It always helps to answer the questions that are actually asked.
Sorry to the OP. Thanks to Mr. H!

So is this the equation I use for part c and d, but I have to combine the mass of both the bucket and glass?
 
  • #36
Xazerd said:
So is this the equation I use for part c and d, but I have to combine the mass of both the bucket and glass?

Not for part c)

You got part c.

Gravity the sole actor for the centripetal force in c, so mass cancels.

Part d) would occur at the bottom of the circular path. So the tension, or more directly the normal force would act up, and gravity acts down. If you combine mass of the glass and the bucket then you can use tension.
 
  • #37
pgardn said:
Not for part c)

You got part c.

Gravity the sole actor for the centripetal force in c, so mass cancels.

Part d) would occur at the bottom of the circular path. So the tension, or more directly the normal force would act up, and gravity acts down. If you combine mass of the glass and the bucket then you can use tension.

Ok. Got it. Thanks!
 
  • #38
Xazerd said:
Ok. Got it. Thanks!

Sorry I did not read the question more carefully and used part a) for everything.
 
  • #39
No Problem!
 

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